optimasi

1. The Theory of Interest - Solutions Manual
23
Chapter 3
1. The equation of value using a comparison date at time 20t = is
20 10
50,000 1000 at 7%.s Xs= +
Thus,
20
10
50,000 1000 50,000 40,995.49
$651.72.
13.81645
s
X
s
− −
= = =
2. The down payment (D) plus the amount of the loan (L) must equal the total price paid
for the automobile. The monthly rate of interest is .18/12 .015j = = and the amount
of the loan (L) is the present value of the payments, i.e.
( )48 .015
250 250 34.04255 8510.64.L a= = =
Thus, the down payment needed will be
10,000 8510.64 $1489.36.D = − =
3. The monthly interest rate on the first loan (L1) is 1 .06/12 .005j = = and
( )( )1 48 .005
500 500 42.58032 21,290.16.L a= = =
The monthly interest rate on the second loan (L2) is 2 .075/12 .00625j = = and
2 125,000 25,000 21,290.16 3709.84.L L= − = − =
The payment on the second loan (R) can be determined from
12 .00625
3709.84 Ra=
giving
3709.84
$321.86.
11.52639
R = =
4. A’s loan: 8 .085
20,000 Ra=
20,000
3546.61
5.639183
R = =
so that the total interest would be
( )( )8 3546.61 20,000 8372.88.− =
B’s loan: The annual interest is
( )( ).085 20,000 1700.00=
so that the total interest would be
( )( )8 1700.00 13,600.00.=
Thus, the difference is
13,600.00 8372.88 $5227.12.− =

2. The Theory of Interest - Solutions Manual Chapter 3
24
5. Using formula (3.2), the present value is
( )1 1 1
where .
n
n
n i
na i
i n
−
⎡ ⎤− +⎣ ⎦= =
This expression then becomes
2
1
1
1 .
1 1
n
n
n
n
nn
n
n
n
−
⎡ ⎤+⎛ ⎞
−⎢ ⎥⎜ ⎟ ⎡ ⎤⎛ ⎞⎝ ⎠⎣ ⎦ = −⎢ ⎥⎜ ⎟
+⎝ ⎠⎣ ⎦
6. We are given
1
,
n
n
v
a x
i
−
= = so that 1 .n
v ix= − Also, we are given
2
2
1
,
n
n
v
a y
i
−
= = so that 2
1 .n
v iy= − But ( )22n n
v v= so that ( )2
1 1 .iy ix− = − This
equation is the quadratic ( )2 2
2 0x i x y i− − = so that 2
2
.
x y
i
x
−
= Then applying
formula (1.15a), we have 2
2
.
1 2
i x y
d
i x x y
−
= =
+ + −
7. We know that 1 ,d v= − and directly applying formula (3.8), we have
( ) ( )8 88
8
1 1 1 1 .9
5.695.
.1
v d
a
d d
− − − −
= = = =
8. The semiannual interest rate is .06/ 2 .03.j = = The present value of the payments is
( ) ( )21 9
100 100 15.87747 8.01969 $2389.72.a a+ = + =
9. We will use a comparison date at the point where the interest rate changes. The
equation of value at age 65 is
25 .08 15 .07
3000s Ra=
so that
25 .08
15 .07
3000 236,863.25
$24,305
9.74547
s
R
a
= = =
to the nearest dollar.
10. (a) Using formulas (3.1) and (3.7)
( )
( )
2 1
2
1
1 1 .
n n n
n
n n n
n
a v v v v v
v v v v a v
−
= + + + + + −
= + + + + − = + −

3. The Theory of Interest - Solutions Manual Chapter 3
25
(b) Using formulas (3.3) and (3.9)
( ) ( ) ( )
( ) ( ) ( )
( )
1
1
1 1 1 1 1
1 1 1 1 1
1 1 .
n n
n
n n
n
n
s i i i
i i i
s i
−
−
⎡ ⎤= + + + + + + + −⎣ ⎦
⎡ ⎤= + + + + + + + −⎣ ⎦
= − + +
(c) Each formula can be explained from the above derivations by putting the
annuity-immediate payments on a time diagram and adjusting the beginning and
end of the series of payments to turn each into an annuity-due.
11. We know that
( )1 11
and .
qp
p q
iv
a x s y
d i
+ −−
= = = =
Thus, 1 1p
v dx ivx= − = − and ( )1 1 ,
q
i iy+ = + so that ( )
1
1 .q
v iy
−
= +
Finally,
( ) ( )
( )
1 1 1
1
1
1 1
.
1 1
p q
p q
v ivx
a
i i iy
iy ivx vx y
i iy iy
+
+
− −⎛ ⎞
= = −⎜ ⎟+⎝ ⎠
+ − − +
= =
+ +
12. We will call September 7, 1 0z t− =
so that March 7, 8 is 34z t+ =
and June 7, 12 is 51z t+ =
where time t is measured in quarters. Payments are made at 3t = through 49,t =
inclusive. The quarterly rate of interest is .06/ 4 .015.j = =
(a) ( ) ( )49 2
PV 100 100 34.5247 1.9559 $3256.88.a a= − = − =
(b) ( ) ( )32 15
CV 100 100 40.6883 13.3432 $5403.15.s a= + = + =
(c) ( ) ( )49 2
AV 100 100 71.6087 2.0150 $6959.37.s s= − = − =
13. One approach is to sum the geometric progression
( )
45
15 30 45
1515 15 15 45
15
1
1 .
1
av
a v v a a a
v a
−
+ + = = =
−
The formula also can be derived by observing that
( )15 30
15 3015 15 15 15 45
1a v v a a a a+ + = + + =
by splitting the 45 payments into 3 sets of 15 payments each.

4. The Theory of Interest - Solutions Manual Chapter 3
26
14. We multiply numerator and denominator by ( )
4
1 i+ to change the comparison date
from time 0t = to 4t = and obtain
( )
( )
4
7 7 3 4
4
11 7 411
1
.
1
a a i a s
a a sa i
+ +
= =
++
Therefore 4, 7, and 4.x y z= = =
15. The present value of annuities X and Y are:
( )
10
30 10
20
10 10
PV and
PV .
X
Y
a v a
K a v a
= +
= +
We are given that PV PVX Y= and 10
.5.v = Multiplying through by i, we have
( ) ( )( )30 10 10 10 20
1 1 1 1v v v K v v− + − = − +
so that
10 20 30
10 20 30
1 1 .5 .25 .125 1.125
1.8.
1 1 .5 .25 .125 .625
v v v
K
v v v
+ − − + − −
= = = =
− + − − + −
16. We are given 5 1010 5
3a a= ⋅ or 5 10
10 5
3v a v a= and
10 5
5 101 1
3 .
v v
v v
i i
− −
=
Therefore, we have
5 15 10 15
3 3v v v v− = − or 15 10 5
2 3 0v v v− + = or ( ) ( )
5 10
2 3 1 1 0i i− + + + =
which is a quadratic in ( )
5
1 .i+ Solving the quadratic
( )
( ) ( )( )( )2
5 3 3 4 2 1 3 1
1 2
2 2
i
± − − ±
+ = = =
rejecting the root 0.i =
17. The semiannual interest rate is .09/ 2 .045.j = = The present value of the annuity on
October 1 of the prior year is 10
2000 .a Thus, the present value on January 1 is
( )
( )( )( )
.5
10
2000 1.045
2000 7.91272 1.02225 $16,178
a
= =
to the nearest dollar.
18. The equation of value at time 0t = is
30
20
1000a R v a∞
= ⋅ ⋅
or
20
301 1
1000
v
R v
d d
−
= ⋅

5. The Theory of Interest - Solutions Manual Chapter 3
27
so that
( )( )
( ) ( )
20
3020
30
30 10
1
1000 1000 1 1
1000 1 1 .
v
R v i
v
i i
−
= = − +
⎡ ⎤= + − +⎣ ⎦
19. We are given
1
9
i = so that
1
.
1 10
i
d
i
= =
+
The equation of value at time 0t = is
( ) ( )1 1 .1
6561 1000 or 6.561 .
.1
n n
n d
v a
d∞
− −
= = =
Therefore, ( ) ( )( ).9 .1 6.561 .6561
n
= = and 4.n =
20. The equation of value at age 60 is
5
20
50,000a Rv a∞
=
or
20
550,000 1 v
Rv
i i
−
=
so that
5 25
50,000
at .05
50,000
$102,412
.7835262 .2953028
R i
v v
= =
−
= =
−
to the nearest dollar.
21. Per dollar of annuity payment, we have PV PVA D= which gives
1
or 3
3
n n
n n
a v a a v a∞ ∞
= ⋅ =
and 1 3n n
v v− = , so that
( )4 1 or .25 and 1 4.
nn n
v v i= = + =
22. Per dollar of annuity payment, we have
2 3
PV , PV , PV and PV .n n n
A B C Dn n n
a v a v a v a∞
= = = =
We are given
2PV
.49 or .7.
PV
n nC
A
v v= = =

6. The Theory of Interest - Solutions Manual Chapter 3
28
Finally,
( )
( )
3 3
22
PV 1
PV
1 1 .7 .30 30
.
.49 49.7
n n n
nB
n n
D
n
n
v a v v
v a v
v
v
∞
−
= =
− −
= = = =
23. (a)
( )
( )
( )
.25
5.25
5.25 5
.25
1 1
at .05
1.05 1
4.32946 .77402 4.5195.
.05
i
a a v i
i
⎡ ⎤+ −
= + =⎢ ⎥
⎣ ⎦
⎡ ⎤−
= + =⎢ ⎥
⎣ ⎦
(b)
( )( )
5.25
5.25 5
.25
4.32946 .25 .77402 4.5230.
a a v= +
= + =
(c)
( )( )
6
5.25 5
.25
4.23946 .25 .74621 4.5160.
a a v= +
= + =
24. At time 0t = we have the equation of value
( )4
4
1000 100 or
10 13.5875 at .045.
n
n
a a
a a i
= −
= + = =
Now using a financial calculator, we find that 21n = full payments plus a balloon
payment. We now use time 21t = as the comparison date to obtain
( )21
17
1000 1.045 100s K= +
or
( )
( )
21
17
1000 1.045 100
2520.2412 100 24.74171 46.07
K s= −
= − =
Thus, the balloon payment is
100 46.07 $146.07 at time 21.t+ = =
25. We are given 1 2PV PV= where
1 236 18
PV 4 and PV 5 .a a= =
We are also given that ( )1 2.
n
i+ = Thus, we have
( ) ( )( ) ( )
36 18
36 18 18 18
1 1
4 5 or
4 1 4 1 1 5 1 .
v v
i i
v v v v
− −
⋅ =
− = − + = −

7. The Theory of Interest - Solutions Manual Chapter 3
29
Thus, we have
( )18 18
4 1 5 or .25.v v+ = =
Finally, we have ( )
18
1 4,i+ = so that ( )
9
1 2i+ = which gives 9.n =
26. At time 20,t = the fund balance would be
20
500 24,711.46 at .08.s i= =
Let n be the number of years full withdrawals of 1000 can be made, so that the
equation of value is
1000 24,711.46 or 24.71146n n
s s= = .
Using a financial calculator we find that only 14n = full withdrawals are possible.
27. (a) The monthly rate of interest is .12/12 .01.j = = The equation of value at time
0t = is
( )
( )
60
6000 100 4495.5038
ln .749251
.749251 so that 29.
ln 1.01
k
k
v a
v k
= =
−
= = =
(b) Applying formula (2.2) we have
( )
( )
( )( )
( )
1000 1 2 60 60 61 61
30.5.
100 60 2 60 2
t
+ + +
= = = =
28.(a) Set: N 48 PV 12,000 PMT 300= = = − and CPT I to obtain .7701%.j =
The answer is 12 9.24%.j =
(b) We have 48
300 12,000a = or 48
40.a = Applying formula (3.21) with 48n =
and 40,g = we have
( )
( )
( )
( )
2 2 48 40
.8163%.
1 40 48 1
n g
j
g n
− −
≈ = =
+ +
The answer is 12 9.80%.j =
29. We have
( ) ( )1 22
2
or 1.75 1 1 .a v v i i
− −
= + = + + +
Multiplying through ( )2
1 i+ gives
( ) ( )
( )
2
2
1.75 1 1 1
1.75 1 2 2
i i
i i i
+ = + +
+ + = +
and 2 2
1.75 2.5 .25 or 7 10 1 0i i+ − + − = which is a quadratic. Solving for i

8. The Theory of Interest - Solutions Manual Chapter 3
30
( ) ( )( )( )
( )( )
2
10 10 4 7 1 10 128
2 7 14
4 2 5
rejecting the negative root.
7
i
− ± − − − ±
= =
−
=
30. We have the following equation of value
10 20
10,000 1538 1072 .a a= =
Thus ( ) ( ) ( )( )10 20 10 10
1538 1 1072 1 1072 1 1 ,v v v v− = − = − + so that 10 1538
1
1072
v+ = or
10
.43470.v =
Solving for i, we obtain
( ) ( )10 .1
1 .43470 and .43470 1 .0869, or 8.69%.i i
− −
+ = = − =
31. We are given that the following present values are equal
7.25% 50 1
.j n j
a a a∞ −
= =
Using the financial calculator
50
1
13.7931
.0725j
a = =
and solving we obtain 7.00%.j = Since 1 6%,j − = we use the financial calculator
again
6%
13.7931n
a = to obtain 30.2.n =
32. (a) We have 1 .08/ 2 .04j = = and 2 .07/ 2 .035.j = = The present value is
( ) ( )( )6
6 .04 4 .035
1.04 5.2421 3.6731 .79031
8.145.
a a
−
+ = +
=
(b) The present value is
( ) ( )( )6
6 .04 4 .035
1.035 5.2421 3.6731 .81350
8.230.
a a
−
+ = +
=
(c) Answer (b) is greater than answer (a) since the last four payments are discounted
over the first three years at a lower interest rate.
33. (a) Using formula (3.24)
( ) ( ) ( ) ( )
2 3 4 5
5
2 3 4 5
1 1 1 1 1
1.06 1.062 1.064 1.066 1.068
4.1543.
a v v v v v= + + + +
= + + + +
=

9. The Theory of Interest - Solutions Manual Chapter 3
31
(b) Using formula (3.23)
( )( ) ( )( )( ) ( )( )( )( )
( )( )( )( )( )
5
1 1 1 1
1.06 1.06 1.062 1.06 1.062 1.064 1.06 1.062 1.064 1.066
1
4.1831.
1.06 1.062 1.064 1.066 1.068
a = + + +
+ =
34. Payments are R at time .5t = and 2R at time 1.5, 2.5, , 9.5.t = … The present value of
these payments is equal to P. Thus, we have
( ) ( )
1
24
4
1 2 2 1 1i a j
P R a a i i
− −
⎡ ⎤= + + + +
⎣ ⎦
and
( )
( )
1
2
4
4 5
1
.
1 2 2 1i j
P i
R
a i a
−
+
=
+ + +
35. The payments occur at 0, 1, 2, , 19t = … and we need the current value at time 2t =
using the variable effective rate of interest given. The current value is
1 1 1
1 1 1
1 1 1 1 1 1
1 1 1 1 1 1 1
9 10 10 11 11 12
1 1 1
1 1 1
11 12 27
10 11 11 11 11 12 11 12 27
1
9 10 10 12 12 13 12 13 28
− − −
− − −
⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + + + + + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛
= + + + + + + ⋅⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝
28
9
11 11 11 11 11 11 11
.
9 10 11 12 13 28 t t=
⎞
⎟
⎠
= + + + + + + = ∑
36. We know that ( )1
1a t dt−
= − using simple discount. Therefore, we have
( ) ( ) ( )1 1
2
1 1
1 1
n n
n
t t
a a t dt n n n d−
= =
= = − = − +∑ ∑
by summing the first n positive integers.
37. We have ( )
( ) ( )2 2
2
1 1
,
2log 2 log 1 log
1
a t
tt t
t
= =
++ − +
+
so that ( )1
2
2
log .
1
t
a t
t
− +
=
+

10. The Theory of Interest - Solutions Manual Chapter 3
32
Now
( )
( )
1 1
1
2
0 0
2 2 2
2 2
2
log
1
2 3 1
log log log
1 2
2 3 1
log log 1 .
1 2
n n
n
t t
t
a a t
t
n
n
n
n
n
− −
−
= =
+
= =
+
+
= + + +
+⎛ ⎞
= ⋅ ⋅ ⋅ = +⎜ ⎟
⎝ ⎠
∑ ∑
38. The accumulated value of 1 paid at time t accumulated to time 10 is
( )
1010 1
ln 20 ln1020
20
.
10
r tt
drdr rr
r
e e e
δ − −−
−∫∫ = = =
Then
10
10
1
20 19 18 10
14.5.
10 10 10 10r
r
s
=
−
= = + + + =∑
39.
1 1 1 1 1
A: PV 4.8553
1.01 1.02 1.03 1.04 1.05
B: AV 1.04 1.03 1.02 1.01 1.00 5.1000
A
B
= + + + + =
= + + + + =
and taking the present value
5.1000
PV 4.8571.
1.05
B = =
The answers differ by 4.8571 - 4.8553=.0018.
40. The present value of the payments in (ii) is
( )10 20 10 20
10 10 10 10
30 60 90 30 60 90 .a v a v a a v v+ + = + +
The present value of the payments in (i) is
( )10
20 10
55 55 1 .a a v= +
Equating the two values we have the quadratic 20 10
90 5 25 0.v v+ − = Solving the
quadratic
( ) ( )( )( )
( )( )
2
10 5 5 4 90 25 90
.5
2 90 180
v
− ± − −
= = =
rejecting the negative root. Now 10
.5v = or ( )
10
1 2i+ = and .0718.i = Finally,
20 .0718
55 574.60.X a= =
41. We have the equation of value at time 3t n=
3 2
98 98 8000n n
s s+ =
or

11. The Theory of Interest - Solutions Manual Chapter 3
33
( ) ( )
3 2
1 1 1 1 8000
81.6327.
98
n n
i i
i i
+ − + −
+ = =
We are given that ( )1 2.
n
i+ = Therefore,
3 2
2 1 2 1 10
81.6327
i i i
− −
+ = = and .1225,i =
or 12.25%.
42. At time 0t = we have the equation of value
20 15 10 5
10,000 4ka ka ka ka= − − −
so that
20 15 10 5
10,000
.
4
k
a a a a
=
− − −
43. The present values given are:
(i) 2
2 36n n
a a+ = or ( ) ( )2
2 1 1 36 ,n n
v v i− + − = and
(ii) 2 6n
n
v a = or ( )2 1 6 .n n
v v i− =
Thus, ( ) ( ) ( )( ) ( )2
2 1 1 6 2 1n n n n
v v v v− + − = − which simplifies to the quadratic
2
10 13 3 0.n n
v v− + =
Solving,
( ) ( )( )( )
( )( )
2
13 13 4 10 3 6
.3
2 10 20
n
v
± − −
= = =
rejecting the root 1.n
v = Substituting back into (ii)
( )( )1 .3
2 .3 6,
i
−
= so that
( )( )( )2 .3 .7
.07,
6
i = = or 7%.
44. An equation of value at time 10t = is
( ) ( )( ) ( )( )
( ) ( )
10 6 5
4 3
10,000 1.04 1.05 1.04 1.05 1.04
1.04 1.04 10,000.
K K
K K
− −
− − =
Thus, we have
( )
( )( ) ( )( ) ( ) ( )
10
6 5 4 3
10,000 1.04 1
1.05 1.04 1.05 1.04 1.04 1.04
$980 to the nearest dollar.
K
⎡ ⎤−⎣ ⎦
=
+ + +
=
45. ( )
40 40
41 15
15 15
261
1 1
n
n
n n
s s
s i
i i= =
− −
⎡ ⎤= + − =⎣ ⎦∑ ∑
using formula (3.3) twice and recognizing that there are 26 terms in the summation.